In a circle with centre 'O' PA and PB are two chords. PC is the chord that bisects the angle . APB The tangent to the circle at C is drawn meeting PA and PB extended at Q and R respectively. If QC = 3, QA = 2 and RC = 4, then length of RB equals

Explanation

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the Geometry
We have a circle with center 'O', chords PA and PB, and a chord PC that bisects the angle ∠APB. A tangent at point C meets the extensions of PA and PB at points Q and R, respectively. We are given the lengths QC = 3, QA = 2, and RC = 4, and we need to find the length of RB.

### Step 2: Use the Power of a Point Theorem
According to the Power of a Point theorem, for point P, we have:
\[ PQ \cdot PA = QC^2 \]
Let PQ = x. Then:
\[ x \cdot 2 = 3^2 \]
\[ x \cdot 2 = 9 \]
\[ x = \frac{9}{2} \]

### Step 3: Find PR Using Similar Triangles
Using the sine rule in triangle PQR, we can relate the segments:
\[ \frac{QC}{PQ} = \frac{RC}{PR} \]
Substituting the known values:
\[ \frac{3}{\frac{9}{2}} = \frac{4}{PR} \]
Cross-multiplying gives:
\[ 3 \cdot PR = 4 \cdot \frac{9}{2} \]
\[ 3 \cdot PR = 18 \]
\[ PR = 6 \]

### Step 4: Use the Power of a Point Again
Now, we apply the Power of a Point theorem again for triangle PCR:
\[ PR \cdot RB = RC^2 \]
Substituting the known values:
\[ 6 \cdot RB = 4^2 \]
\[ 6 \cdot RB = 16 \]
\[ RB = \frac{16}{6} = \frac{8}{3} \]

### Conclusion
Thus, the length of RB is:
\[ \boxed{\frac{8}{3}} \]

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